MathsCBSE Class 10 · Standard · Sample paper, session 2026-27 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard sample paper 2026-27: question 34(A)

From the official CBSE sample question paper for 2026-27, section D, question 34(A), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34(A) · 5 marks · Long answerCompetency-basedSurface Areas and Volumes

A tent is in the shape of a cylinder surmounted by a conical top. If the height and radius of the cylindrical part are 3 m and 14 m respectively, and the total height of the tent is 13.5 m, then find the area of the canvas required for making the tent, keeping a provision of $26 m^{2}$ of canvas for stitching and wastage. Also, find the cost of the canvas to be purchased at the rate of $₹ 250$ per $m^{2}$ .

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

Open one hint at a time and try again before the next.

Hint 1

Only the curved surfaces are covered by canvas: the cylinder's side and the cone's side.

Hint 2

Cone height = total height − cylinder height; then use $l = r^{2} + h^{2}$ .

Hint 3

Total canvas = $π r l + 2 π r h + 26$ ; then multiply by ₹250.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Figure (in words): a cylinder of radius $r = 14$ m and height $3$ m, with a cone of the same radius on top. The total height is $13.5$ m.
Height of the conical part: $h = 13.5 - 3 = 10.5$ m
½ mark
Slant height of the cone: $l = r^{2} + h^{2} = 1 4^{2} + 10. 5^{2} = 196 + 110.25 = 306.25 = 17.5$ m
1 mark
Curved surface area of the cone: $π r l = \frac{22}{7} \times 14 \times 17.5 = 770 m^{2}$
1 mark
Curved surface area of the cylinder: $2 π r h = 2 \times \frac{22}{7} \times 14 \times 3 = 264 m^{2}$
The base of the tent is open ground, so no canvas is needed for it.
1 mark
Canvas required = CSA of cone + CSA of cylinder + wastage
$= 770 + 264 + 26 = 1060 m^{2}$
½ mark
Cost of canvas $= 1060 \times 250 = ₹ 2, 65, 000$
1 mark

Answer: Canvas required = 1060 m²; cost = ₹ 2,65,000

Where the marks are

Finds height of cone = 10.5 m (½ mark)
Finds slant height = 17.5 m (1 mark)
Finds CSA of cone = 770 m² (1 mark)
Finds CSA of cylinder = 264 m² (1 mark)
Adds wastage to get total canvas = 1060 m² (½ mark)
Finds cost ₹ 265000 (1 mark)

Revise the chapter

Chapter: Surface Areas and Volumes (notes, formulas, where marks are lost and every past question)
Topic: Surface Area and Volume Formulae Revision

Practise this chapterBoard-style questions from Surface Areas and Volumes, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2026-27 (Standard): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Standard sample paper 2026-27: question 34(A)

Hints

Step-by-step solution

Revise the chapter

More questions from Surface Areas and Volumes